The ball's horizontal and vertical velocities at time [tex]t[/tex] are
[tex]v_x=v_{xi}[/tex]
[tex]v_y=v_{yi}-gt[/tex]
but the ball is thrown horizontally, so [tex]v_{yi}=0[/tex]. Its horizontal and vertical positions at time [tex]t[/tex] are
[tex]x=v_{xi}t[/tex]
[tex]y=20.0\,\mathrm m-\dfrac g2t^2[/tex]
The ball travels 22 m horizontally from where it was thrown, so
[tex]22\,\mathrm m=v_{xi}t[/tex]
from which we find the time it takes for the ball to land on the ground is
[tex]t=\dfrac{22\,\rm m}{v_{xi}}[/tex]
When it lands, [tex]y=0[/tex] and
[tex]0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2[/tex]
[tex]\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}[/tex]
